If $P$ is any point on the ellipse $16x^2 + 25y^2 = 400$ with foci $S$ and $S'$, and the area of $\Delta PSS'$ is $9 \text{ square units}$, then the abscissa of point $P$ is:

$A$ vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $PQR$,$\Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$,then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

The mid-point of a chord of the ellipse $x^2+4y^2-2x+20y=0$ is $(2,-4)$. The equation of the chord is

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle SBS^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

If $\alpha$ and $\beta$ are the eccentric angles of the endpoints of a focal chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2} = ....$